Block A is released from rest from wedge B. find out velocity of the wedge 'C' when the bloack A slide down completely after climbing up on it. All surfces are smooth. Block A, wedge B and wedge C are at rest initially.

Options

A. `sqrt(2gH)`
B. `sqrt(gH)`
C. `sqrt((16gh)/(27))`
D. `sqrt((4gH)/(9))`

Explanation

When A slide down from the wedge B by conservatio of momentum and energy
`0=mv_(A)-2mv_(B)`
`(1)/(2)2mv_(B)^(2)+(1)/(2)mv_(A)^(2)=mgH`
When block A climbs up and then climbs down from C.
By conservation of momentum and energy velocity of with `C=(2m)/(m+2m)V_(A)`